Question

Find the zeroes of the polynomial p(x) =x2-x-12

Answer 1

Answer:

x=4 and x=-3 are the zeroes of the polynomial

Step-by-step explanation:

$p(x) = {x}^{2} - x - 12 = 0 \\ \\ {x}^{2} - 4x + 3x - 12 = 0 \\ \\ x(x - 4) + 3(x - 4) = 0 \\ \\ (x - 4)(x + 3) = 0 \\ \\ x - 4 = 0 \\ x = 4 \\ \\ x + 3 = 0 \\ x = - 3$

Answer 2

Answer:

HOPE THIS HELPS

Step-by-step explanation:

x^2-x-12=0

x^2-4x+3x-12=0

x(x-4)+3(x-4)=0

(x+3)(x-4)=0

x+3=0      x-4=0

x=-3          x=4